When a horse pulls a 56 kg buggy down the road with a force of 73 N, the buggy hitch makes an angle of 25 degrees above the horizontal. If the force of friction on the buggy is 5 N, what is the buggy's acceleration? Report your answer to 2 decimal places.
To find the buggy's acceleration, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In equation form, it can be written as:
ΣF = m * a
where ΣF is the net force, m is the mass of the object, and a is the acceleration.
First, we need to calculate the net force acting on the buggy. The net force is the vector sum of all the forces acting on the buggy. In this case, the forces acting on the buggy are the force applied by the horse and the force of friction.
To calculate the force applied by the horse, we need to resolve the force into its x and y components. The x-component of the force is given by:
F_x = F * cos(theta)
where F is the magnitude of the force and theta is the angle with respect to the horizontal. Plugging in the values, we have:
F_x = 73 N * cos(25°)
Next, we need to calculate the y-component of the force, which is given by:
F_y = F * sin(theta)
Plugging in the values, we have:
F_y = 73 N * sin(25°)
Now, let's calculate the net force:
ΣF = F_x - F_friction
where F_friction is the force of friction. Plugging in the values, we have:
ΣF = F_x - 5 N
Once we have the net force, we can use it with the mass of the buggy to find the acceleration. Rearranging the equation ΣF = m * a, we get:
a = ΣF / m
Plugging in the values, we have:
a = (ΣF) / (56 kg)
Finally, we can calculate the acceleration:
a = (F_x - F_friction) / (56 kg)
Substituting the values we calculated earlier, we have:
a = (73 N * cos(25°) - 5 N) / (56 kg)
Evaluating the expression, we find the buggy's acceleration to be approximately 1.07 m/s².